1. The easiest way is to arrange the eighteen matches as in Diagrams 1
and 2, making the length of the perpendicular AB equal to a match and a
half. Then, if the matches are an inch in length, Fig. 1 contains two
square inches and Fig. 2 contains six square inches--4 × 1½. The second
case (2) is a little more difficult to solve. The solution is given in
Figs. 3 and 4. For the purpose of construction, place matches
temporarily on the dotted lines. Then it will be seen that as 3 contains
five equal equilateral triangles and 4 contains fifteen similar
triangles, one figure is three times as large as the other, and exactly
eighteen matches are used.
[Illustration: Figures 1, 2, 3, 4.]